7
$\begingroup$

Update (January 2018) A new very interesting paper with various experiments on the IBM machine is Five Experimental Tests on the 5-Qubit IBM Quantum Computer by Diego García-Martín, Germán Sierra.

Update (July 2017) Extended question: Of course, the question extends to the newly built IBM 16-qubits quantum computer, and the 50-qubits quantum computers that IBM, Google and various other players expect to demonstrate in the near future. A natural conjecture is that all these computers can only approximate "noise-stable" pure states. namely pure states that are stable under noise operator with exponential decay of high order Pauli terms, (see e.g. this paper by Montanaro and Osborne).

Original question:

IBM has recently built a 5-qubits quantum computers based on superconducting qubits. It is even possible to make experiments over the cloud.

The space of pure states for 5 qubits is the unit sphere in the 32-dimensional space over the complex number. When you measure the state of the computer you get a probability distribution on strings of 5 bits.

Give a "map" of the probability distributions and of the actual states that the IBM quantum computer can reach.

Of course, the boundary between what can be reached and what is beyond reach is of special interest. And also the case of 3 and 4 qubits.

My question is motivated by this comment over Lipton abd Reagan's blog by Alexander Vlasov.

I would expect that distributions that can actually be reached are characterized by "noise stability" which is equivalent to "low degree Fourier-type" expansion. The relevant "noise" here is independent depolarizing noise. The relevant "Fourier expansion" might be the one described by Ashley Montanaro and Tobias J. Osborne in the paper quantum Boolean function.

Vlasov referred to the paper by Daniel Alsina and José Ignacio Latorre Experimental test of Mermin inequalities on a 5-qubit quantum computer. A sentence from the abstract reads "The experimental results obtained using the quantum computer show violation of all Mermin inequalitites, with a clear degradation of the results in the 5 qubit case."

Update: Andrew Morgan suggests in the comment below that the proposal for the threshold might be too naive. A specific question inspired by Andrew's comment is:

Can we well approximate with the IBM machine the state

$$\frac 14 |11111\rangle+\frac 14 |11100\rangle+ \frac 14 |11010\rangle+\frac 14 |11001\rangle+ \frac 14 |10110\rangle+\frac 14 |10101\rangle+ \frac 14 |01110\rangle+\frac 14 |01101\rangle+ \frac 14 |10011\rangle+\frac 14 |01011\rangle+ \frac 14 |00111\rangle+\frac 14 |10000\rangle+ \frac 14 |01000\rangle+\frac 14 |00100\rangle+ \frac 14 |00010\rangle+\frac 14 |00001\rangle.$$ (Superposition of the computational basis elements with an odd numbers of 1's)

(Later update: In view of Peter Shor's comment below I suppose we need to look more carefully for "noise-sensitive" states that we would like to explore.)

Also of interest are the states $$\frac {1}{\sqrt {5}} |10000\rangle + \frac {1}{\sqrt {5}} |01000\rangle +\frac {1}{\sqrt {5}} |00100\rangle + \frac {1}{\sqrt {5}} |00010\rangle +\frac {1}{\sqrt {5}} |00001\rangle $$ and the superposition of all basis elements with two 1's. (If somebody reading this question actually tried to program the IBM 5-qubit computers I'd love to hear about it.)

$\endgroup$
  • 1
    $\begingroup$ In what sense might qubit-efficient universal computation fit into this? I'm thinking of the following: Imagine for a moment that that any boolean function on $n$ bits can be computed using only the output qubit and no ancillary bits as $|x b\rangle \mapsto |x(b\oplus f(x))\rangle$. Then one might compute the uniform distribution over all even-weight strings by preparing a uniform superposition in the first $n-1$ bits and $|0\rangle$ as the last qubit, and then computing parity of the first $n-1$ bits into the last bit. Am I understanding right that this would disagree with your conjecture? $\endgroup$ – Andrew Morgan Jun 8 '16 at 21:42
  • 2
    $\begingroup$ The state you chose is a stabilizer state (stabilized by $-ZZZZZ$, $XXIII$, $IXXII$, $IIXXI$, $IIIXX$). Did you chose it on purpose ? Even if a quantum machine could produce and sample all stabilizer states, it could not prove any supremacy, because of the Gottesmann-Knill theorem. $\endgroup$ – Frédéric Grosshans Jun 10 '16 at 15:35
  • 4
    $\begingroup$ I hope you realize that the state you wrote down, in a different basis, is $(|+++++\rangle - |-----\rangle)/\sqrt{2}$, where $|+\rangle = (|0\rangle +|1\rangle)/\sqrt{2}$ and $|-\rangle = (|0\rangle -|1\rangle)/\sqrt{2}$. $\endgroup$ – Peter Shor Jun 11 '16 at 19:36
  • 2
    $\begingroup$ Gil: ctually, in standard quantum theory, that is a relatively noise-sensitive state. However, maybe you think there are states which are much more noise-sensitive in your theory of decoherence. I'd like to see what they are and why they're more noise-sensitive. $\endgroup$ – Peter Shor Jun 13 '16 at 18:16
  • 1
    $\begingroup$ In terms of density matrices, this is a 1023-dimensional space. I don't think stating it in terms of probability distributions makes it substantially easier. For arbitrarily many qubits, a generalization of this problem is probably unsolvable. This is not an easy question. $\endgroup$ – Peter Shor Jun 19 '16 at 14:13
7
$\begingroup$

I just ran the first state you suggest (i.e. the GHZ state with negative phase in the Hadamard basis). Basically what I did was to write a circuit which creates that state, apply one of 5 stabilizer generators (XIXII,IXXII,IIXXI,IIXIX or -ZZZZZ) and then inverted the creation process. The circuit is shown below for the -ZZZZZ stabilizer.

enter image description here

There seems to be problem with the calibration or something, since I am seeing a bit flip occurring on the first qubit with roughly probability 2/3 that should not happen. The expected result should be 00000 in all cases, however 10000 seems to be twice as likely in almost all cases (prob ~0.1 vs ~0.2 over 1024 runs for each stabilizer generator).

Below are the probabilities of obtaining 00000 for each stabilizer generator over 1024 runs each: XIXII:0.10840 IXXII:0.15625 IIXXI:0.11621 IIXIX:0.10645 ZZZZZ:0.09375

And the probabilities of obtaining 10000 for each stabilizer generator: XIXII:0.21777 IXXII:0.21875 IIXXI:0.17871 IIXIX:0.14648 ZZZZZ:0.19726

$\endgroup$
  • $\begingroup$ @GilKalai: If you want the full statistics, just drop me an email. They are too long to post here. $\endgroup$ – Joe Fitzsimons Jun 20 '16 at 19:14
  • $\begingroup$ you could post it in a pastebin and then link it. $\endgroup$ – Joshua Herman Jan 14 '18 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.